Monday, May 9, 2016

Welcome to The Lazy Chemist!

The purpose of this site is to provide some basic info and explanation to fundamental techniques used in organic chemistry lab. Still a work in progress, but for now, check out the published extraction and NMR pages.

Sunday, May 8, 2016

Extraction Flowchart Example

You may have an entire lab dedicated to separating organic compounds using extraction methods, but you will likely be using extraction in numerous labs throughout the year. The lab dedicated to extraction often has the students separating two or three organic compounds from each other by using only extraction techniques. In this post, I'm going to walk through the flowchart for separating three compounds: 3-methylphenol, 2-ethyl-3-methylbenzoic acid, and ethylbenzene.

Let's assume we have these molecules in a single solution of dichloromethane. Our goal is to separate them by extraction.

Eachof these molecules have aromatic rings (hydrophobic) but two of them also have ionizable groups (OH for 3-methylphenol and COOH for 2-ethyl-3-methylbenzoic acid). By using acid/base reactions, you can make these molecules charged/ionized which makes them more polar and therefore able to partition to the aqueous phase.

The following is a flowchart designed to separate these three molecules. In this example, the steps are already predetermined, the task is for the student to identify the structures of the molecules at each step:

In the first step, treating with a weak base (sodium bicarbonate) will convert the COOH to a carboxylate salt (COO- Na+) which is more water-soluble:

RCOOH  +  NaHCO3    <---->    RCOO-  +   Na+    +     H2CO3

By doing so, the benzoic acid molecule would become negatively charged (this makes it very polar), and because of this, it becomes much more soluble in water and will partition to the aqueous phase.

The 3-methylphenol has an OH group, but alcohols typically require strong base to deprotonate the H and become charged. Therefore, treatment with sodium bicarbonate (weak base) will have no effect on 3-methylphenol, and it will remain in the dichloromethane phase (organic layer).

Ethylbenzene has no ionizable groups, so treatment with any base or acid will have no effect on ethylbenzene. 

After the first step, the flowchart would now look like this:

In your aqueous layer, you would have your 2-ethyl-3-methyl benzoic acid molecule, but as the carboxylate ion form. In order to get it back to its carboxylic acid form, you just have to add acid:

RCOO-  +   Na+   +   HCl   <----->    RCOOH  +  NaCl

In your organic layer, you would still have a mixture of 3-methylphenol and ethylbenzene. Since 3-methylphenol has an ionizable OH group, you can treat this with a strong base:

ROH  +  NaOH   <----->   RO-   +   Na+    +   H2O

By doing so, the molecule becomes negatively charged (polar) which makes it able to partition to the aqueous phase. Ethylbenzene is not affected by NaOH, so would remain hydrophobic and remain in the organic layer. 

The flowchart would now look like this:

At this point, we have successfully isolated 2-ethyl-3-methylbenzoic acid as well as ethylbenzene.

In the last step, we just have to acidify the ionic form of 3-methylphenol. Like before, we just need to treat this with acid:

RO-  +  Na+   +   HCl    <------>     ROH    +    NaCl

The flowchart is now complete:

We started out with a mixture of three different organic molecules in a single solution of dichloromethane. There are numerous separation techniques that organic chemists use to isolate their molecules of interest, but extraction is one of the simplest approaches because it is generally quick and only requires a separatory funnel. 

By taking advantage of the ionizable groups of the molecules here, we were easily able to manipulate their water-solubility by treating with acid/bases accordingly. Ultimately, we were successfully able to design a extraction protocol to isolate all three components!

Saturday, April 23, 2016


Common questions about extraction:

  • What is extraction?
  • Why do we do extraction?
  • What's the difference between extraction and "washing"
  • What's the organic phase?
  • What's the aqueous phase?
  • Why is the organic phase sometimes on top and other times on bottom?
  • How do we know which layer is the organic phase???
Extraction is an important skill that is learned early on because you end up using it in nearly every lab afterward. It is a very fast and (often) efficient way to separate a compound of interest from a mixture, and it only requires solvent and a separation funnel.

What is extraction?

The term "extraction" in this context means that your compound of interest is initially in one solvent and then partitions to a different, immiscible solvent (the extraction solvent). For example, you have caffeine dissolved in water, and you want to remove it from the water. To do this, you can mix with dichloromethane (immiscible organic solvent). By mixing, a fraction of the caffeine will partition into the dichloromethane solvent because it is more soluble in dichloromethane than water. Two solvent layers will then form (water and dichloromethane), and the majority of the caffeine will now be in the dichloromethane.

Why would you want to do this? Often in a case like this, there would be other impurities present in the initial water layer, and our goal would be to purify the caffeine from these other impurities. It doesn't get much simpler than mixing another solvent in and shaking. Also, another common reason for an extraction like this is to have your compound of interest in a more volatile solvent that is easy to remove. If your caffeine is in water, it'll take a VERY long time to remove the water and have an idea of how much caffeine you have. Instead, you can extract it into dichloromethane which has a low boiling point, and you can remove the solvent by roto-evaporation in minutes.

What's the difference between extraction and washing?

Sometimes you will see in your procedure that you need to perform "extraction" and other times it says you need to perform "washing," but both are using the same types of solvents and using a separatory funnel, so it sounds like it's the exact same thing. I never knew there was an actual difference until someone dumbed it down for me in grad school.

Simply, extraction is moving your compound of interest from one solvent to another. Washing is keeping your compound of interest in the same solvent, but the second solvent is used to "extract" other impurities.

Using the same caffeine example before, let's say you have your caffeine in your dichloromethane solvent, but you know there are traces of water droplets and salt still present in your sample. The easiest way to clean this up is to add some water and separate the two layers. This would be called "washing" because your compound of interest (caffeine) remains in the same solvent layer (dichloromethane). The impurities are actually the ones getting extracted because they are going from dichloromethane to the water layer.

What's the organic phase?

In every case (in orgo lab at least), the organic phase is an organic solvent that is NOT miscible with water. Common organic phase solvents used in the lab are diethyl ether, ethyl acetate, dichloromethane (aka methylene chloride), chloroform, hexanes, and benzene.

What's the aqueous phase?

The aqueous phase is the water-based phase that is NOT miscible with the organic phase. So obviously water would be an aqueous phase, but this also includes most acids, bases, and salt solutions. So this would include any time you have 10% HCl, 10% NaOH, acetic acid, sodium chloride, brine, sodium bicarbonate, etc.

Why is the organic phase sometimes on top and other times on bottom?

This is one of my favorite questions, and fortunately the answer is so simple! It's simply based on the density of the organic solvent. If the organic solvent is MORE dense than water, the organic phase will be on the bottom. If the organic solvent is LESS dense than water, it will be on top. It doesn't matter much if the aqueous phase contains HCl, NaOH, or some salt, the density of the aqueous phase will be roughly 1 g/mL. So for example, if you extract with chloroform or dichloromethane, your organic phase will be on bottom. If you extract with diethyl ether or ethyl acetate, your organic phase will be on top.

How do we know which layer is the organic phase???

If you're in the lab and don't have quick access to density values for your solvents (this is why we have you do pre-lab reagent tables!!), then you can do the water drop test. First, your layers must be separated, so either you have two layers in your sep funnel, or you've actually collected them separately into two separate flasks. Let's say they are both in your sep funnel. Simply add a drop or two of water from the top, and see where the water droplets go. If they seem to disappear in the top layer, then your top layer is the aqueous. If you see your water droplets fall down to the bottom layer, then your bottom layer is aqueous. 

Tips for lazy chemists:

  • Make a flowchart if there are multiple extraction/wash steps and understand which components will be in the aqueous or organic phase. It can get confusing and overwhelming very quickly when you're doing it in the lab.
  • Don't discard anything until the lab is finished. You don't want to accidentally throw away your product during one of the steps.
  • When in doubt about which phase is which, do the water drop test! 
  • For a practice flowchart problem, check out our example here.

Saturday, March 19, 2016

How To Approach NMR Problems

There is no perfect way to approach an NMR problem. Often students don't even know where to start. I'd like to offer my thoughts on how to approach figuring out a structure when given an NMR spectrum.

There are 3 main things you must consider (not necessarily in this order):
chemical shift (ppm value)
integration (how many H that make up the peak)
peak shape (singlet, doublet, triplet, etc)

Chemical shift
The ppm value of a peak gives you many hints as to what sort of H group it is. 
Insert graph or figure
Look for the obvious peaks such as aromatics, aldehydes, and carboxylic acids

This tells you how many H's make up a particular peak. This doesn't tell us what type of H it is, but it tells us how many, and this will help us figure out the overall structure of the molecule.

Peak shape
The shape of the proton peak tells us how many protons are on adjacent carbons. Remember the n+1 rule? 

How to approach a problem
Now that we understand the three most important characteristics of the spectrum, we can start to figure out the structure of the molecule. Again, there's no perfect way to do this. But I suggest using a fragmentation approach. In other words, figure out the fragments of the molecule first, and then try to piece them together and see if it all agrees with your spectrum. Let's work through a problem so you see what I mean.

We'll start with a simple one. Often you are given other data, such as IR, mass spec, molecular formula, or at least the proton integration. We'll make it easy for this first one and provide the molecular formula which is C4H8O.

The integration of the peak at 2.5 ppm is 2H
The integration of the peak at 2.2 ppm is 3H
The integration of the peak at 1.1 ppm is 3H

How do we solve it? So first off, there's no aromatic protons, no aldehyde peaks no broad singlets (so no alcohols), so we can rule out all these groups. Can't be carboxylic acid or ester either because these require 2 oxygens. Since there is 1 oxygen based on the molecular formula, there may be an ether or ketone.

Let's look at peak shape. The first one that stands out is the singlet at 2.2 ppm. This means that it's "isolated" (in other words this group is not adjacent to a carbon that has C-H bonds). It also has an integration of 3H. So most likely, this is an isolated CH3 group.

Let's look at the peak at 1.1 ppm. It is a triplet which means it neighbors a carbon (or carbons) that is bound to a total of 2 protons. This peak also has an integration of 3H, so it is likely a CH3 group. Since it's a triplet, it must neighbor CH groups on either side, or it neighbors a CH2 group on one side. 

The peak at 2.5 ppm is a quartet (neighbors carbon(s) bound to 3 protons total). The peak has an integration of 2H, so likely a CH2 group. We saw previously that we have a CH3 group that could perhaps neighbor a CH2 group, so perhaps this is the CH2 group that neighbors the CH3 group.

So what do we know so far? We likely have a CH3 and CH2 groups that neighbor each other and then an isolated CH3 group:


So we've accounted for 3 out of the 4 carbons and 8 out of the 8 hydrogens. This means that the 4th carbon is bound to oxygen. Since we've already accounted for all the hydrogens, we can assume that this last carbon isn't bound to any hydrogens, so it must be bound to oxygen through a double bond (so it's a ketone).


Does the structure make sense? Work through the peak shape and integration data to confirm that this structure is consistent with the spectrum (you will find that it is).

What about the chemical shifts? In this case, I didn't find the chemical shifts too valuable or necessary to solve the structure. This will not always be the case, and others may find that the chemical shift data makes more sense to them, so let's discuss briefly. 

Remember that for any given proton peak, its chemical shift value is dependent on the electron density of the adjacent groups. The closer the proton peak to electron dense groups, the higher the ppm value. For example, the CH3 on the left of the structure is adjacent to CH2 (not electron dense), so this group has the lowest ppm value (1.1 ppm). Compare this to the other CH3 group that's on the right of the ketone, it is adjacent to a C=O group which is fairly electron dense due to the double bond and oxygen. Because of this, the CH3 group on the right is shifted more DOWNFIELD (higher ppm value, in this case 2.2 ppm). The CH2 group is the farthest downfield (highest ppm value) because it's a secondary carbon and it also neighbors the ketone. 

With a more complex structure, the chemical shift data becomes very important because it tells us how to put all the fragments together. 

More complicated examples to come!